EE5120 Linear Algebra: Tutorial 7, July-Dec Covers sec 5.3 (only powers of a matrix part), 5.5,5.6 of GS

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1 EE5 Linear Algebra: Tutorial 7, July-Dec 7-8 Covers sec 5. (only powers of a matrix part), 5.5,5. of GS. Prove that the eigenvectors corresponding to different eigenvalues are orthonormal for unitary matrices. Hint: Use properties of unitary matrices See pg. 87 of GS [ b. Suppose A =, where b and c are some non-zero finite real numbers with c =. c [ A O Compute the eigenvalues and eigenvectors of A. Further, if X =, which is a O A block diagonal matrix and O is an all-zero matrix, then what are its eigenvalues and eigenvectors? Hint: Use basic procedure for finding eigenvalues and eigenvectors. A is upper triangular. So, diagonal entries are its eigenvalues, i.e., λ A =, c. Eigenvectors can be obtained as, [ T and [ b c T respectively. X is block diagonal. Since A is upper triangular, so is X. Thus eigen values of X are: λ X =,, c, c and corresponding eigenvectors are [ T, [ T b, [ c T and [ b c T respectively.. Let A be an n n hermitian matrix with n distinct positive eigen values and for any column vector x, let x(k) denote the k th entry in x. (a) Prove that if A = UΛU is the eigen decomposition of A, then (i) entries of Λ are real, and (ii) U is a unitary matrix. Hint: First prove that (i) y H Ay is real for any y, (ii) columns of U can be orthonormal. (b) Let u = u aau, v i = U H u i, i =, and a is a real positive number. If v (k) < v (k), k, then prove that a < λ max, where λ max is the maximum eigen value of A. Hint: Replace A in terms of Λ and U, and then proceed. (a) For any non-zero vector y of appropriate dimension, note that y H Ay is a scalar. Now, (y H Ay) = (y H ) A y = (y T A y ) T = y H A H y = y H Ay, where the third equality is due to the fact that y H Ay is a scalar. Since (y H Ay) = y H Ay, y H Ay is a real number. Now, let p be a eigenvector corresponding to eigenvalue λ for A. Then, p H Ap is real by above arguement. And, p H Ap = p H λp = λ p. When LHS is real and p is real, λ must also be real. Hence, all eigen values of a hermitian matrix

2 are real. Further, let λ and λ be two eigen values of A with eigenvectors p and p respectively. Now, since λ and λ are real and distinct, we have, λ p H p = (λ p ) H p = (Ap ) H p = p H AH p = p H Ap = λ p H p. Since eigenvalues are distinct the above relation holds only if p Hp ph p p p =. Thus, eigen vectors are orthonormal, which implies U is unitary. (b) Now, u = u aau = u auλu H u = (UU H auλu H )u = U(I aλ)u H u. Thus, U H u = (I aλ)u H u v = (I aλ)v. Let k th diagonal entry in Λ be λ k. Then, v (k) = aλ k v (k), k. If we want v (k) < v (k), k, then we must have aλ k <, k < aλ k < < a < λ k. Since, this is true for all k, we have < a < λ max, where λ max is the maximum eighen value of A. 4. Find a third column so that U is unitary. How much freedom in column? / i/ x U = / y i/ / z Hint: Use the definition of unitary matrix. Given U H U = I. On equating LHS and RHS, we get 9 equations of which 4 are redundant. The other 5 equations are given below, From equations (), () and (), x = exp (iθ x ), y = From euqations (4) and (5), θ x θ y = (n + )π and θ x θ z = (k + + x x = () + yȳ = () + + z z = () + xȳ = (4) i + i + x z = (5) exp (iθ y) and z = exp (iθ z ), respectively. y = exp i(θ x (n + )π) = and z = exp i(θ x (k )π) = )π, respectively. So, x = exp (iθ x ), exp i(θ x π) exp i(θ x + π ). Page

3 The third column is exp (iθ x ) exp ( iπ), which has no degree of freedom. exp (i π ) 5. Suppose each Gibonacci number G [ k+ is the average of the two previous numbers G k+ and G k. Then G k+ = (G Gk+ k+ + G k ) : = [ A [ G k+ G k+ G k (a) Find the eigenvalues and eigenvectors of A. Hint: Use basic procedure for finding eigenvalues and eigenvectors. (b) Find the limit as n of the matrices A n = SΛ n S. Hint: Compute Λ. (c) If G = and G =, show that the Gibonacci numbers approach. [ G Hint: Compute A G. (a) A = (b) S = (c) [.5.5. Eigenvalues (λ, s.t. det(a λi) = ) are and -.5, with eigen- vectors (v i, s.t. Av i = λv i ) are [ A = S [ G G 5 5 [.5, S = S = [ 5 [ [ = A G = A G [ and [ 5. =. [ [ 5 5. Suppose there is an epidemic in which every month half of those who are well become sick, and a quarter of those who are sick become dead. Find the steady state for the corresponding Markov process: d k+ s k+ = w k+ 4 4 d k s k w k Hint: Steady state is the eigen vector corresponding eigen value. Let us denote the given matrix equation as u k+ = Au k d k+ where u k+ = s k+, A = w k+ 4 4 and u k = d k s k w k. Page

4 The seady state is given by Au = u, i.e., u is the eigenvector corresponding to eigenvalue (A Markov matrix will always have one of the eigenvalues as and the rest of magnitude ). To find it, we equate (A I)u =. 4 4 d s w = d This gives u = s =, in other words, everyone will die from the epidemic. w 7. Show that every number is an eigenvalue for the transformation T( f (x)) = d f dx, but the transformation T( f (x)) = x f (t)dt has no eigenvalues (here < x < ). Hint: Take f (x) = e ax for any real number a and solve. If f (x) = e ax for any real number a, for the first T, T( f (x)) = d f dx = aeax = a f (x) Hence, any number a is an eigenvalue of T. [Note: the eigenvalues of a linear transformation are not dependent on a basis, so if we have found them in one basis that is good enough. Why? Any change of basis is given by a similarity transform, and it is easy to prove that similar matrices have the same eigenvalues. In this case we chose f (x) = e ax because it was an easy to guess eigenfunction. Since the question is regarding eigenvalues, it is important to only choose those f that are eigenfunctions, not arbitrary functions. For the second T, suppose T( f (x)) = a f (x) for some number a and some function f, i.e., x f (t)dt = a f (x) Now, differentiate both sides to get f (x) = a f (x) Solving for f, we get f (x)dx f (x) = a dx This gives ln f (x) = x a + C or f (x) = e x a +C or f (x) = e C e x a We can get rid of the absolute value signs by substituting A for e C (allowing A to possibly be negative): Page 4

5 f (x) = Ae x a Therefore we have T( f (x)) = x f (t)dt = x Ae t a dt = aae t a x o = aae x a aa = a(ae x a A) = a( f (x) A) But our initial assumption was that T( f (x) = a f (x). For this either a = or A =, either of which implies f (x) =. Hence T has no eigenvalues. 8. Let A be a square matrix. Define B = (A + AH ), B = i (A AH ) (a) Prove that B, B are hermitian and A=B +ib (b) Suppose that A = C +ic, where C and C are hermitian, then prove that C =B and C =B (c) What conditions of B and B make A normal? Hint: Use definitions. (a) B H = (AH + A) = B similarly, B H = i (AH A) = B and B +ib = (A + A H )+ (A AH ) = A (b) A = C +ic, A H = C H +ic H = C - ic using the exressions of A and A H, we get C = (A + AH ) = B and C = i (A AH ) = B (c) AA H =A H A (B +ib )(B -ib )=(B -ib )(B +ib ) simplification yields B B =B B 9. Let A be a normal matrix. Prove the following: (a) Ax = A H x for every x C n (b) A ci is a normal operator for every c C (c) If x is an eigen vector of A with eigen value λ, then x is also an eigen vector of A H with eigen value λ (λ is the complex conjugate of λ ) Hint: Use part (a) and part (b) to solve part (c) (a) Ax =(Ax) H Ax = x H A H Ax = x H AA H x = (A H x) H A H x = A H x (b) (A ci)(a ci) H = AA H c A ca H + c I = A H A c A ca H + c I = (A ci) H (A ci) therefore normal (c) Let x be an eigen vector of A with eigen value λ Ax = λx. Define U = A λi. From part b, U is normal. From part a, Ux = U H x = (A λi)x = A H x λ x A H x = λ x Page 5

6 . If the transformation T is a reflection across the 45 line in the plane, find its matrix with respect to the standard basis v = (, ), v = (, ), and also with respect to V = (, ), V = (, ). Show that those matrices are similar. Hint: Use Change of basis or Similarity transformation B = M AM. Let T be the Transformation matrix for reflection across any θ line given by, T = [ c cs cs s Here, θ = 45. Substituting this value, we get the transformation [ matrix wrt standard basis v, v. Lets call this matrix A, where A =. Now, we change the basis to V = (, ), V = (, ) and find the reflection transformation matrix (call it B). Observe where the new basis vectors land after reflection through the given line, TV = V TV = V [ Thus, the transformation matrix wrt V, V will be B =. Next, to show that A and B are similar matrices, we need to find a matrix M such that, B = M AM. Matrix M is constructed by writing new basis vectors V, V as a linear combination of standard basis vectors v, v. Since (, ) = (, ) + (, ) and (, ) = (, ) (, ), we get [ M = Using this M, it can be verified that B = M AM. Note that, (, ) and (, ) are also the eigenvectors of A (or T). Page